Question

If $(x)^y = (y)^x$, then find $\frac{dy}{dx}$.

Hint:

When differentiating implicit functions involving logarithms, always apply the chain rule and product rule carefully to each term.

Answer 1

We are given the equation: \[ x^y = y^x. \] Taking the natural logarithm of both sides: \[ \log(x^y) = \log(y^x). \] Using the logarithmic identity $\log(a^b) = b \log(a)$, we get: \[ y \log x = x \log y. \] Now, differentiate both sides with respect to $x$. On the left-hand side, apply the product rule to $y \log x$, and on the right-hand side, apply the product rule to $x \log y$: \[ \frac{d}{dx}(y \log x) = \frac{d}{dx}(x \log y). \] Differentiating: \[ \frac{dy}{dx} \log x + y \frac{1}{x} = \frac{dy}{dx} x \frac{1}{y} + \log y. \] Now, collect terms involving $\frac{dy}{dx}$ on one side: \[ \frac{dy}{dx} \log x - \frac{dy}{dx} \frac{x}{y} = \log y - \frac{y}{x}. \] Factor out $\frac{dy}{dx}$: \[ \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x}. \] Finally, solve for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{\log y - \frac{y}{x}}{\log x - \frac{x}{y}}. \]